POJ 2947 Widget Factory(高斯消元)

题目连接～～～
 　　本题是对同余方程组消元，方程比较好列，本题可以列m个方程，每个方程可以列成k1*x1 + k2*x2……kn*xn = y1 (mod 7)
 y1是加工这一批零件所需要的时间。然后解方程就可以了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <ctime>
#include <algorithm>
using namespace std ;
#define INF 0x3f3f3f3f
const double esp = 0.00000001 ;
const int MX = 300 + 10 ;
const int MY = 100 + 10 ;

int g[MX][MX] ,ans[MX] ;
int Solve(char s[]){
    int f_num = 0 ;
    if(strcmp(s ,"MON") == 0) f_num = 1 ;
    else if(strcmp(s ,"TUE") == 0) f_num = 2 ;
    else if(strcmp(s ,"WED") == 0) f_num = 3 ;
    else if(strcmp(s ,"THU") == 0) f_num = 4 ;
    else if(strcmp(s ,"FRI") == 0) f_num = 5 ;
    else if(strcmp(s ,"SAT") == 0) f_num = 6 ;
    else if(strcmp(s ,"SUN") == 0) f_num = 7 ;
    return f_num ;
}
int Gcd(int a ,int b){
    int tmp ;
    while(b){
        tmp = a ;
        a = b ;
        b = tmp % b ;
    }
    return a ;
}
int Lcm(int a ,int b){
    return a/Gcd(a ,b)*b ;
}
int Gauss(int n ,int m ,int g[][MX] ,int mod){//n行 m列
    int max_r ;
    int k = 0 ,col = 0 ;
    for( ;k < n && col < m ; ++k ,++col){
        max_r = k ;
        for(int i = k+1 ;i < n ; ++i)
            if(abs(g[max_r][col])  < abs(g[i][col]))
                max_r = i ;
        if(!g[max_r][col]){
            k-- ; continue ;
        }
        if(max_r != k){
            for(int i = col ;i <= m ; ++i)
                swap(g[k][i] ,g[max_r][i]) ;
        }
        for(int i = k + 1 ;i < n ; ++i){
            if(g[i][col]){
                int LCM = Lcm(abs(g[k][col]) ,abs(g[i][col])) ;
                int ta = LCM/abs(g[i][col]) ;
                int tb = LCM/abs(g[k][col]) ;
                if(g[i][col] * g[k][col] < 0) tb = -tb ;
                for(int j = col ;j <= m ; ++j)
                    g[i][j] = ((g[i][j]*ta - g[k][j]*tb)%mod +mod)%mod ;
            }
        }
    }
    for(int i = k ;i < n ; ++i)
        if(g[i][col]) return -1 ;
    if(k < m) return m-k ;
    for(int i = n-1 ;i >= 0 ; --i){
        int tmp = g[i][m] ;
        for(int j = i+1 ;j < m ; ++j){
            tmp -= g[i][j]*ans[j] ;
            tmp = (tmp%mod + mod)%mod ;
        }
        int sum ;
        for(int t = 3 ;t <= 9 ; ++t){
            sum = ((g[i][i]*t)%mod + mod)%mod ;
            if(sum == tmp){
                ans[i] = t ;
                break ;
            }
        }
    }
    return 0 ;
}
int main(){
    //freopen("input.txt" ,"r" ,stdin) ;
    char s1[MX] ,s2[MX] ;
    int n ,m ,mod = 7 ,num ,tmp ;
    while(~scanf("%d%d" ,&n ,&m)){
        if(!n && !m) break ;
        memset(g ,0 ,sizeof(g)) ;
        for(int i = 0 ;i < m ; ++i){
            scanf("%d%s%s" ,&num ,s1 ,s2) ;
            int d1 = Solve(s1) ;
            int d2 = Solve(s2) ;
            g[i][n] = ((d2 - d1 + mod)%mod + 1)%mod ;
            for(int j = 0 ;j < num ; ++j){
                scanf("%d" ,&tmp) ;
                g[i][tmp-1] = (g[i][tmp-1]+1)%mod ;
            }
        }
        memset(ans ,0 ,sizeof(ans)) ;
        int f_num = Gauss(m ,n ,g ,mod) ;
        if(!f_num) {
            for(int i = 0 ;i < n ; ++i)
                if(!i) printf("%d" ,ans[i]) ;
                else printf(" %d" ,ans[i]) ;
            puts("") ;
        }
        else if(f_num == -1) puts("Inconsistent data.") ;
        else puts("Multiple solutions.") ;
    }
    return 0 ;
}